Question

If the range of the function $f(x)=\frac{x-1}{p-x^2+1}$ does not contain any values belonging to the interval $\left[-1,\frac{-1}{3}\right]$ then the true set of values of $p$, is

• A. $(-\infty ,-1)$
• B. $\left(-\infty ,\frac{-1}{4}\right)$
• C. $(0,\infty )$
• D. $(-\infty ,0)$

Answer 1

Answer: (B)

We have $y=\frac{x-1}{p-x^2+1}\Rightarrow py-x^{2}y+y=x-1\Rightarrow x^{2}y+x-y(p+1)-1=0$
As, $x\in R$, so $D\ge 0\Rightarrow 1+4y(y(p+1)+1)\ge 0\Rightarrow 4y^2(p+1)+4y+1\ge 0$
Since $y\ne \left[-1,\frac{-1}{3}\right]$
So, $4y^2(p+1)+4y+1<0\forall y\in \left[-1,\frac{-1}{3}\right]$
$\Rightarrow (2y+1{)}^2+4y^{2}p<0$
$\Rightarrow p<-1{\left(\frac{2y+1}{2y}\right)}^2\forall y\in \left[-1,\frac{-1}{3}\right]$
Hence $p<-{{\left(\frac{2y+1}{2y}\right)}^2|}_{\min }$
Now max. value of ${\left(\frac{2y+1}{2y}\right)}^2$ occurs at $y=-1$ and is equal to $\frac{-1}{4}$.
$\therefore p<\frac{-1}{4}\Rightarrow \text{ B }$